Question 77: Gravitation

Question

An astronaut takes a ball of mass $$m$$ from earth to space. He throws the ball into a circular orbit about earth at an altitude of $$318.5 \mathrm{~km}$$. From earth's surface to the orbit, the change in total mechanical energy of the ball is $$x \frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{21 \mathrm{R}_{\mathrm{e}}}$$. The value of $$x$$ is (take $$\mathrm{R}_{\mathrm{e}}=6370 \mathrm{~km})$$ :

Accepted Answer

At earth surface, $$E_1=-\frac{G M_m}{R_e}$$
in the orbit, $$E_2=-\frac{G M_m}{2 r}$$
$$\begin{aligned}
\Delta E & =G M_m\left[\frac{1}{R_e}-\frac{1}{2 r}ight] \\
& =G M_m\left[\frac{1}{R_e}-\frac{1}{2.1 R_e}ight] \\
& =\frac{11}{21} \frac{G M_m}{R_e} \\
\Rightarrow & x=11
\end{aligned}$$

Answer

12

Answer

11

Answer

9

Answer

10

Gravitation

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